Radius of Gyration Class 11
As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we define a new parameter known as the radius of gyration.
It is the distance whose square when multiplied with the mass of the body gives the moment of inertia of the body about the given axis. For a body of mass $M,$ if $K$ is radius of gyration
$\mathrm{I}=\mathrm{MK}^{2}$
$\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}} \quad \mathrm{K}=\sqrt{\frac{\sum_{i=1}^{n} m_{i} r_{i}^{2}}{M}}$
Through this concept a real body (particular) is replaced by a point mass for dealing its rotational motion. e.g., in case of a solid sphere rotating about an axis through its centre of mass.
$\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{(2 / 5) \mathrm{MR}^{2}}{\mathrm{M}}}=\mathrm{R} \sqrt{\frac{2}{5}}$
So instead of solid sphere we can assume a point mass M at a distance $(\mathrm{R} \sqrt{\frac{2}{5}})$ from the axis of rotation for dealing the rotational motion of the solid sphere.
K depends on
(i) axis of rotation(ii) distribution of mass of body
K does not depend on
(i) Mass of the body $\quad$
(ii) Position of body
(iii) On all angular physical quantity
- radius of gyration does not depends on the mass of rigid body. Graphical relation between log K and log I.
Relation between angular momentum and angular velocity
$J=I \omega$ I- Moment of inertia with respect to axis of rotation $\omega$ - Angular velocity due to angular momentum J- The moment of momentum which is causing rotational motion. The rate of change of angular momentum is equal to the torque applied on the body. $$ \vec{\tau}=\frac{d \vec{J}}{d t} $$ In rotational motion angular momentum has equal importance as linear momentum in linear motion If torque acting of a particle is zero then $\vec{\tau}=0 \Rightarrow \frac{d \vec{J}}{d t}=0 \quad$ or $\quad \vec{J}=$ constant Which implies that the angular momentum remians conserved when no external torque acts on the body. LAW OF CONSERVATION OF ANGULAR MOMENTUM If no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant that is $\vec{J}=\overrightarrow{\mathrm{I} \omega}=\mathrm{constant}$ Proof: For a rigid body, rotating about a given axis. $\quad \vec{\tau}=1 \vec{\alpha}=\mathrm{I} \cdot \frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}$ Now, for a rigid body, rotating about a given axis, Iremains constant. $\vec{\tau}=\mathrm{I} \cdot \frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}=\frac{\mathrm{d}(\overrightarrow{\mathrm{I} \omega})}{\mathrm{dt}}=\frac{\mathrm{d} \overrightarrow{\mathrm{J}}}{\mathrm{dt}}$ If the external torque acting on the body, $\vec{\tau}=0$ $$ \frac{d \overrightarrow{\mathrm{J}}}{d t}=0, \text { and hence } \overrightarrow{\mathrm{J}}=\text { constant. } $$ For the equation $\mathrm{J}=\mathrm{I} \omega,$ it can be seen that if $\mathrm{J}$ remains constant, $\omega$ increases when I decreases and vice versa. If I changed to $\mathrm{I}_{1}$, then, in the absence of an external torque, $\omega$ will change to $\omega_{1}$, such that $\mathrm{I}_{\mathrm{i}} \omega_{\mathrm{i}}=\mathrm{I}_{\mathrm{f}} \omega_{\mathrm{f}}$ This is the law of conservation of angular momentum. Introduction to Rotational Dynamics Moment of Inertia Moment of Inertia: Perpendicular and Parallel axis theorem Radius of Gyration Law of Conservation of Angular Momentum Conservation of Angular Momentum Examples Kinetic Energy of a Rotating Body Work done in rotatory Motion Rotational Power Combine Translational and Rotational Motion Rolling without slipping Rolling on a plane surface Rolling on a Inclined Plane For Latest updates, Subscribe our Youtube ChannelClick here to get exam-ready with eSaral
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